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Math Help - Finding common ratio and more?

  1. #1
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    Finding common ratio and more?

    I actually don't even know what topic this is. O.o"
    It's a three part problem and I will probably ask lots of questions because I don't think I've learned this yet.

    An infinite geometric series whose terms are positive has a finite sum. The ratio of the sum of the first four terms to the sum of the first two terms is 13:9.
    a) Calculate the value of the common ratio r.

    b) If the fourth term is 16/9, calculate the sum of the infinite series.

    c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.
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  2. #2
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    Quote Originally Posted by Jukixel View Post
    I actually don't even know what topic this is. O.o"
    It's a three part problem and I will probably ask lots of questions because I don't think I've learned this yet.

    An infinite geometric series whose terms are positive has a finite sum. The ratio of the sum of the first four terms to the sum of the first two terms is 13:9.
    a) Calculate the value of the common ratio r.

    b) If the fourth term is 16/9, calculate the sum of the infinite series.

    c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.
    Hints:

    Let the series be  S = a + a r + a r^2 +a r^3 + a r^4 + \cdots. You said the series has a finite answer so it converges and thus  -1 < r < 1

    For (a) consider \frac{a + ar + ar^2 + ar^3}{a + ar} = \frac{13}{9}

    which factors

    \frac{a(1+r)(1+r^2)}{a(1+r)} = \frac{13}{9} from which you can find r.

    Once you know r you should be able to find a (that's using the fact in (b)). Do you know the formula for the sum of a converging infinite geometric series?
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    No, don't think I know it.
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  4. #4
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    Quote Originally Posted by Jukixel View Post
    No, don't think I know it.
    If  -1 < r < 1 then the sum of

    S = a + ar + ar^2 + ar^3 + \cdots

    is

    S_{\infty} = \frac{a}{1-r}
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    The ratio of the sum of the first four terms to the sum of the first two terms is 13:9
    a) Calculate the value of the common ratio r.
    let r be the common ratio ...

    \frac{a + ar + ar^2 + ar^3}{a + ar} = \frac{13}{9}

    \frac{1 + r + r^2 + r^3}{1 + r} = \frac{13}{9}

    9 + 9r + 9r^2 + 9r^3 = 13 + 13r

    9r^3 + 9r^2 - 4r - 4 = 0

    9r^2(r + 1) - 4(r + 1) = 0

    (r+1)(9r^2 - 4) = 0

    (r+1)(3r+2)(3r-2) = 0

    r = -1 or r = -\frac{2}{3} or r = \frac{2}{3}

    An infinite geometric series whose terms are positive has a finite sum.
    r = \frac{2}{3} is the only valid solution

    b) If the fourth term is 16/9, calculate the sum of the infinite series.
    ar^3 = \frac{16}{9}

    sum of a finite infinite geometric series is S = \frac{a}{1-r}


    c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.
    sum of the first n terms of a geometric series is S_n = \frac{a(1-r^n)}{1-r}
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    Thanks!
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